### Electric Field Intensity (E) Due To a Circular Ring Charge - Field Theory.

- Consider a circular ring of radius 'a' which carries a uniform line charge density ρL as shown in figure.

- We need to find out electric field at a point P (0, 0, h) on the z axis (z > 0).

- Electric field intensity (E) due to any line charge (ρL) in general is given as:

In this case,

dl = a dφ (Since the differential part dl is a differential arc)

R2 = a2 + h2

Consider the triangle shown in the above figure

a aρ + R = h az → R = - a aρ + h az

aR = R / | R |

aR = - a aρ + h az / ( a2 + h2 )1/2

Substituting all these values in the above equations, the electric field intensity E becomes:

- For every element dl there is a corresponding element diametrically opposite that gives an equal but opposite dEρ so that the two contributions cancel each other.

Hence contribution along aρ due to symmetry adds up to zero.

- Therefore the final electric field intensity at point (0, 0, h) has only z component.

- Hence Electric field intensity (E) due to a circular ring (of radius a carrying a uniform charge ρL) placed on the x-y plane and if the point of interest is any point on z axis, then it is given as:

- Similarly if the ring is placed on x-z plane and point of interest is any point on y- axis, then E is given as:

- Introduction To Electrostatics.

- Coulomb's law.

- Electric Field Intensity (E).

- Electric Lines Of Forces /Streamlines / Electric Flux (ψ) .

- Electric Flux Density (D).

- Electric Field Intensity Due To a Finite and Infinite Line Charge.

- Electric Field Intensity Due To a Infinite Sheet Charge.

- Electric Field Intensity Due To a Circular Ring Charge.

- Electric Field Intensity Due To a Circular Disk Charge.

- Numericals / Solved Examples - Electric Force and Field Intensity.

- Numericals / Solved Examples - Electric Field Intensity - Line, Surface and Mixed Charge Configuration.

Short Notes/FAQ's

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2. Proof for the above derivation

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